(36.) If through the point of bisection of the base of a triangle any line be drawn, intersecting one side of the triangle, and the other produced, and meeting a parallel to the base from the vertex; this line will be cut harmonically. From the vertex B of the E triangle ABC, let BE be drawn parallel to the base AC, and through the middle point D let any line EGF be drawn meeting AB, BC, BE, in F, G, E; F FE: FD :: BE : AD, B but BE (DC=) DA :: EG: GD, since the triangles BGE, DGC are equiangular, .. (Eucl. v. 15.) EG: GD :: FE : FD, or the line is divided harmonically. (37.) If from either angle of a triangle a line be drawn intersecting that which joins the vertex and the bisection of the base, the opposite side, and the line from the vertex parallel to the base; it will be cut harmonically. From the vertex A of the triangle ABC, let AE be drawn pa rallel to the base BC, and AD to its point of bisection D; and from C draw any line CFGE; then will CE CF: EG: FG. F H Draw GH parallel to BC. Since AE and BC are parallel, (Eucl. vi. 2.) BA: AG: CE EG, and since GH is parallel to BD, BA: AG: BD: GH:: DC: GH, :: CF : FG, since the triangles DFC, GHF are similar; .. CE: EG :: CF : FG, and CE CF :: EG: FG. (38.) To draw a line from one of the angles at the base of a triangle, so that the part of it cut off by a line drawn from the vertex parallel to the base, may have a given ratio to the part cut off by the opposite side. From Alet AE be drawn parallel to BC. Divide AB in G, so that AB AG in the given ratio; join CG, and produce it to meet AE in E. CGE is the line required. E B For the triangles AGE, BGC are equiangular, whence (Eucl. v. 18.) CE: EG: BA: AG, i. e. in the given ratio. (39.) To determine that point in the base produced of a right-angled triangle, from which the line drawn to the angle opposite to the base shall have the same ratio to the base produced, which the perpendicular has to the base itself. Let AB be the base, and CB the perpendicular of a right-angled triangle. Draw CE at right angles to AC, meeting AB produced in E. At the point C make the angle ECD=CAB. Dis the point required. From D draw DF perpendicular to AC, and .. parallel to CE. Since the angle FDC is equal to the alternate angle DCE, i. e. to CAB, and the angles at F and B are right angles, .. the triangles DCF, ACB are equiangular; and DAF is also equiangular to ACB, hence FD : DA :: BC: CA, and DC: DF :: AC: AB, .. ex æquo per. CD: DA :: CB : BA. (40.) If the base of any triangle be divided into two parts by a line which is a mean proportional between them, and which being drawn parallel to the second side is terminated in the third; any line parallel to the base will be divided by the mean proportional (produced if necessary) into segments, which will be to each other inversely as the whole mean proportional to that segment which is terminated in the third side of the triangle. Let AC the base of the triangle ABC be divided into two parts in D, by a line DE which is parallel to BC, and a mean proportional between AD and DC; then any line FG parallel to AC, and meeting H H G B DE (produced if necessary) in H, will be divided into segments FH, HG, which are to each other inversely as the lines DE, HE. For since FH is parallel to AD, FH: AD: HE: DE, but AD: DE :: DE : (DC=) HG, (41.) If from the extremities of the base of any triangle, two straight lines be drawn intersecting each other in the perpendicular, and terminating in the opposite sides; straight lines drawn from thence to the intersection of the perpendicular with the base, will make equal angles with the base. From A and C, the extremities of AC, the base of the triangle ABC, let AE, CF be drawn intersecting the perpendicular BD in the same point G. Join FD, ED; these lines make equal angles FDA, EDC with the base. H M Draw EI, FH perpendicular, and KGL parallel to the base; then FH is parallel to BD, and.. BG BD :: FM: FH. And in the same manner it may be shewn that But FM whence FM: FH :: EN : EI; and.. FM EN :: FH: EI. EN :: FG: GN :: KG: GL :: HD: DI, .. HD: DI :: HF: EI, whence the two triangles DFH, DEI, having the angle at H equal to the angle at I, and the sides about the equal angles proportional, are equiangular; .. the angle HDF is equal to EDI. (42.) In every triangle, the intersection of the perpendiculars drawn from the angles to the opposite sides, the intersection of the lines from the angles to the middle of the opposite sides, and the intersection of the perpendiculars from the middle of the sides, are all in the same straight line. And the distances of those points from one another are in a given ratio. From the angles A and B of the triangle ABC, let AD, BE be drawn perpendicular to the opposite sides, H will be the intersection of the three perpendiculars (vii. 34.). From A and B H draw AG, BF to the points of bisection of the opposite sides, intersecting in K, which.. is (iii. 28.) the intersection of the lines drawn from the angles to the middle of the opposite sides; and from F and G draw the perpendiculars FI, GI meeting in I, which.. (iii. 27.) is the intersection of the three perpendiculars. Join HK, KI; HKI is a straight line. Join GF. (Eucl. vi. 2.) AB is parallel to, and double of GF; .. by similar triangles ABK, KFG, BK is double of KF, and AK double of KG. And the triangles AHB, FIG are equiangular, .. AH is double of IG, and BH is double of IF; and .. BH : IF :: 2 : 1 :: BK : KF, whence the triangles BHK, KIF having the angles at Band F equal, and the sides about them proportional, are similar, .. the angle HKB is equal to IKF, :. H, K, and I are in the same straight line. And since BK is double of KF, HK is double of KI, and their distances from each other will be in an invariable ratio. |